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MATHS WORKSHOPSDifferentiation
Business School
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation
bull A useful way to explore the properties of a function is to findthe derivative
Definition (Derivative)
The derivative is a measure of how a function changes as its inputchanges More
bull The derivative of a function at a chosen input value describesthe best linear approximation of the function near that inputvalue
bull For single variable functions f(x) the derivative at a pointequals the slope of the tangent line to the graph of thefunction at that point
bull The process of finding a derivative is called differentiation
Differentiation Rules Application Conclusion
Tangent
Definition (Tangent)
The tangent to a curve at a given point is the straight line thatldquojust touchesrdquo the curve at that point More
Example
x
f(x)
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h1
f(a + h1)
f(a + h1)minus f(a)
h1
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h2
f(a + h2)f(a + h2)minus f(a)
h2
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
hasymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
f prime(a) =∆y
∆x=
rise
run
Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2 by first principles)
Using the definition on the previous slide
d
dxf(x) = f prime(x) = lim
hrarr0
f(x + h)minus f(x)
h
= limhrarr0
(x + h)2 minus x2
h
= limhrarr0
x2 + 2xh + h2 minus x2
h
= limhrarr0
h(2x + h)
h
= limhrarr0
(2x + h)
= 2x
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation
bull A useful way to explore the properties of a function is to findthe derivative
Definition (Derivative)
The derivative is a measure of how a function changes as its inputchanges More
bull The derivative of a function at a chosen input value describesthe best linear approximation of the function near that inputvalue
bull For single variable functions f(x) the derivative at a pointequals the slope of the tangent line to the graph of thefunction at that point
bull The process of finding a derivative is called differentiation
Differentiation Rules Application Conclusion
Tangent
Definition (Tangent)
The tangent to a curve at a given point is the straight line thatldquojust touchesrdquo the curve at that point More
Example
x
f(x)
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h1
f(a + h1)
f(a + h1)minus f(a)
h1
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h2
f(a + h2)f(a + h2)minus f(a)
h2
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
hasymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
f prime(a) =∆y
∆x=
rise
run
Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2 by first principles)
Using the definition on the previous slide
d
dxf(x) = f prime(x) = lim
hrarr0
f(x + h)minus f(x)
h
= limhrarr0
(x + h)2 minus x2
h
= limhrarr0
x2 + 2xh + h2 minus x2
h
= limhrarr0
h(2x + h)
h
= limhrarr0
(2x + h)
= 2x
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation
bull A useful way to explore the properties of a function is to findthe derivative
Definition (Derivative)
The derivative is a measure of how a function changes as its inputchanges More
bull The derivative of a function at a chosen input value describesthe best linear approximation of the function near that inputvalue
bull For single variable functions f(x) the derivative at a pointequals the slope of the tangent line to the graph of thefunction at that point
bull The process of finding a derivative is called differentiation
Differentiation Rules Application Conclusion
Tangent
Definition (Tangent)
The tangent to a curve at a given point is the straight line thatldquojust touchesrdquo the curve at that point More
Example
x
f(x)
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h1
f(a + h1)
f(a + h1)minus f(a)
h1
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h2
f(a + h2)f(a + h2)minus f(a)
h2
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
hasymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
f prime(a) =∆y
∆x=
rise
run
Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2 by first principles)
Using the definition on the previous slide
d
dxf(x) = f prime(x) = lim
hrarr0
f(x + h)minus f(x)
h
= limhrarr0
(x + h)2 minus x2
h
= limhrarr0
x2 + 2xh + h2 minus x2
h
= limhrarr0
h(2x + h)
h
= limhrarr0
(2x + h)
= 2x
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiation
bull A useful way to explore the properties of a function is to findthe derivative
Definition (Derivative)
The derivative is a measure of how a function changes as its inputchanges More
bull The derivative of a function at a chosen input value describesthe best linear approximation of the function near that inputvalue
bull For single variable functions f(x) the derivative at a pointequals the slope of the tangent line to the graph of thefunction at that point
bull The process of finding a derivative is called differentiation
Differentiation Rules Application Conclusion
Tangent
Definition (Tangent)
The tangent to a curve at a given point is the straight line thatldquojust touchesrdquo the curve at that point More
Example
x
f(x)
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h1
f(a + h1)
f(a + h1)minus f(a)
h1
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h2
f(a + h2)f(a + h2)minus f(a)
h2
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
hasymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
f prime(a) =∆y
∆x=
rise
run
Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2 by first principles)
Using the definition on the previous slide
d
dxf(x) = f prime(x) = lim
hrarr0
f(x + h)minus f(x)
h
= limhrarr0
(x + h)2 minus x2
h
= limhrarr0
x2 + 2xh + h2 minus x2
h
= limhrarr0
h(2x + h)
h
= limhrarr0
(2x + h)
= 2x
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Tangent
Definition (Tangent)
The tangent to a curve at a given point is the straight line thatldquojust touchesrdquo the curve at that point More
Example
x
f(x)
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h1
f(a + h1)
f(a + h1)minus f(a)
h1
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h2
f(a + h2)f(a + h2)minus f(a)
h2
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
hasymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
f prime(a) =∆y
∆x=
rise
run
Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2 by first principles)
Using the definition on the previous slide
d
dxf(x) = f prime(x) = lim
hrarr0
f(x + h)minus f(x)
h
= limhrarr0
(x + h)2 minus x2
h
= limhrarr0
x2 + 2xh + h2 minus x2
h
= limhrarr0
h(2x + h)
h
= limhrarr0
(2x + h)
= 2x
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h1
f(a + h1)
f(a + h1)minus f(a)
h1
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h2
f(a + h2)f(a + h2)minus f(a)
h2
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
hasymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
f prime(a) =∆y
∆x=
rise
run
Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2 by first principles)
Using the definition on the previous slide
d
dxf(x) = f prime(x) = lim
hrarr0
f(x + h)minus f(x)
h
= limhrarr0
(x + h)2 minus x2
h
= limhrarr0
x2 + 2xh + h2 minus x2
h
= limhrarr0
h(2x + h)
h
= limhrarr0
(2x + h)
= 2x
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
h
asymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
a + h2
f(a + h2)f(a + h2)minus f(a)
h2
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
hasymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
f prime(a) =∆y
∆x=
rise
run
Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2 by first principles)
Using the definition on the previous slide
d
dxf(x) = f prime(x) = lim
hrarr0
f(x + h)minus f(x)
h
= limhrarr0
(x + h)2 minus x2
h
= limhrarr0
x2 + 2xh + h2 minus x2
h
= limhrarr0
h(2x + h)
h
= limhrarr0
(2x + h)
= 2x
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Formal Statement of Differentiation
Definition (Derivative at the point x = a)
d
dxf(a) = f prime(a) = lim
hrarr0
f(a + h)minus f(a)
hasymp ∆y
∆x=
rise
run
This can be explained graphically
x
f(x)
a
f(a)
f prime(a) =∆y
∆x=
rise
run
Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2 by first principles)
Using the definition on the previous slide
d
dxf(x) = f prime(x) = lim
hrarr0
f(x + h)minus f(x)
h
= limhrarr0
(x + h)2 minus x2
h
= limhrarr0
x2 + 2xh + h2 minus x2
h
= limhrarr0
h(2x + h)
h
= limhrarr0
(2x + h)
= 2x
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Example using the technical definition
Example (Differentiate f(x) = x2 by first principles)
Using the definition on the previous slide
d
dxf(x) = f prime(x) = lim
hrarr0
f(x + h)minus f(x)
h
= limhrarr0
(x + h)2 minus x2
h
= limhrarr0
x2 + 2xh + h2 minus x2
h
= limhrarr0
h(2x + h)
h
= limhrarr0
(2x + h)
= 2x
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Notation
bull Given some function of x y = f(x) the following expressionsare equivalent
dy
dx=
d
dxy =
d
dxf(x) = f prime(x)
bull We readdy
dxas ldquothe derivative of y with respect to xrdquo
bull We can differentiate with respect to whatever variable wersquodlike For example if y is a function of u y = f(u) we candifferentiate y with respect to u
dy
du= f prime(u)
bull We read f prime(x) as f prime x This notation is often used forconvenience when there is no ambiguity about what we aredifferentiating with respect to
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiation in practice
Rarely do people differentiate by ldquofirst principlesrdquo ie using thedefinition Instead we use some simple rules More
Function f(x) = y Derivative f prime(x) = dydx
xn nxnminus1
axn anxnminus1
a (some constant) 0
log(x) xminus1 = 1x
ex ex
Example (f(x) = x2)
Using the above rules
f prime(x) =d
dxf(x) = 2x2minus1 = 2x1 = 2x
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) =
9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
=
18x + 3x2
=
3x(6 + x)
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
More complicated example
Example (f(x) = 2x4 + 5x3)
f prime(x) = 2times 4x4minus1 + 5times 3x3minus1
= 8x3 + 15x2
Example (Your turn f(x) = 9x2 + x3)
f prime(x) = 9times 2x2minus1 + 3x3minus1
= 18x + 3x2
= 3x(6 + x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
bull The exponential function is unique in that it is equal to itsderivative
d
dxex = ex
bull The exponential function is sometimes written as
ef(x) = expf(x)
bull In this definition a function of x f(x) is exponentiated
bull The rule for finding a derivative of this type is
d
dxexpf(x) = f prime(x) expf(x)
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) =
3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) =
3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) =
3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Exponential Functions
Example (g(x) = exp2x)
1 Rewrite as g(x) = expf(x) where f(x) = 2x
2 Find f prime(x) = 2x1minus1 = 2x0 = 2
3 Thus gprime(x) = f prime(x) expf(x) = 2 exp2x
Example (Your turn h(x) = exp3x+ 1)
1 Rewrite as h(x) = expf(x) where f(x) = 3x + 1
2 Find f prime(x) = 3x1minus1 + 0 = 3
3 Thus hprime(x) = f prime(x) expf(x) = 3 exp3x + 1
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
bull As with the exponential function there are some special rulesfor differentiating logarithmic (or log) functions
bull Simple cased
dxlog(x) =
1
x
bull General cased
dxlog(f(x)) =
f prime(x)
f(x)
Example (g(x) = log(2x2 + x))
1 Rewrite as g(x) = log(f(x)) where f(x) = 2x2 + x
2 Find f prime(x) = 4x + 1
3 Thus gprime(x) =f prime(x)
f(x)=
4x + 1
2x2 + x
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) =
10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) =
10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=
x(20minus 9x)
10x2 minus 3x3
=
x(20minus 9x)
x2(10minus 3x)
=
20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Differentiating Logarithmic Functions
Example (Your turn y = log(10x2 minus 3x2))
1 Rewrite as y = logf(x) where f(x) = 10x2 minus 3x3
2 Find f prime(x) = 10times 2xminus 3times 3x2 = 20xminus 9x2 = x(20minus 9x)
3 Thus
dy
dx=
d
dxlog(10x2 minus 3x2) =
f prime(x)
f(x)
=x(20minus 9x)
10x2 minus 3x3
=x(20minus 9x)
x2(10minus 3x)
=20minus 9x
x(10minus 3x)
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Specialised Chain Rule)
Let y = [f(x)]ndy
dx= nf prime(x) [f(x)]nminus1
bull This is a special case of the chain rule More
Example (y = (x2 + 2)3)
Here y = [f(x)]n = (x2 + 2)3 so
bull f(x) = x2 + 2 =rArr f prime(x) =d
dx(x2 + 2) = 2x
bull n = 3
bull dydx = nf prime(x) [f(x)]nminus1 = 3times 2xtimes (x2 + 2)3minus1 = 6x(x2 + 2)2
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Product Rule)
Let y = uv where u and v are functions of x
dy
dx= u middot dv
dx+ v middot du
dx= uvprime + vuprime More
Example (y = x3 log(x))
Here u = x3 v = log(x) vprime =dv
dx=
1
x uprime =
du
dx= 3x2 so
dy
dx= utimes vprime + v times uprime
= x3 times 1
x+ log(x)times 3x2
= x2 + 3x2 log(x)
= x2(1 + 3 log(x))
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) =
x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =
d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n =
2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
=
2times 3x2 times (x3 minus 1)2minus1
=
6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = (x3 minus 1)2)
Think of our function as y = [f(x)]n we have in this particularcase
bull f(x) = x3 minus 1
bull f prime(x) =d
dx(x3 minus 1) = 3x2
bull n = 2
So
dy
dx= nf prime(x) [f(x)]nminus1
= 2times 3x2 times (x3 minus 1)2minus1
= 6x2(x3 minus 1)
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u =
2xminus2
which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx=
minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v =
log(xminus 1)
which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1
So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime =
2xminus21
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=
2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn
Example (Your turn y = 2xminus2 log(xminus 1))
Wersquore differentiating a product so think of the function as y = uv
bull Let u = 2xminus2 which means uprime =du
dx= minus 4xminus3
bull Let v = log(xminus 1) which means vprime =dv
dx=
1
xminus 1So
dy
dx= uvprime + vuprime = 2xminus2
1
xminus 1+ (minus4xminus3) log(xminus 1)
=2
x2(xminus 1)minus 4xminus3 log(xminus 1)
Note that you can always check you differentiation usingWolframAlpha ddx 2x^(-2)log(x-1) More
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Other useful differentiation rules
Definition (Quotient Rule)
Let y =u
vwhere u and v are functions of x then
dy
dx=
v middot dudxminus u middot dv
dxv2
=vuprime minus uvprime
v2More
Example (y =x4
ex)
Let u = x4 and v = ex then uprime =du
dx= 4x3 and vprime =
dv
dx= ex
dy
dx=
vuprime minus uvprime
v2=
ex middot 4x3 minus x4 middot ex
(ex)2=
x3(4minus 1)
ex
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Outline
The theory of differentiation
How differentiation is done in practice
Application Finding Maxima and Minima
Conclusion
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Finding Maxima and Minima
The most common use of differentiation is to find the maximumand minimum values of functions
Key Idea
ldquoStationary pointsrdquo occur when the derivative equals zerof prime(x) = 0 ie the tangent line is a horizontal line More
x
f(x)
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Finding Maxima and Minima
To determine if a stationary point is a maximum minimum orneither we find the second order derivatives
Definition (Second order derivative)
The second order derivative of a function f(x) is found by takingthe derivative of the first order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(d
dxf(x)
)
bull If f primeprime(x) lt 0 the stationary point at x is a maximum
bull If f primeprime(x) gt 0 the stationary point at x is a minimum
bull If f primeprime(x) = 0 the nature of the stationary point must bedetermined by way of other means often by noting a signchange around that point
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3)
We found previously that f prime(x) = 8x3 + 15x2 = x2(8x + 15)
bull To find the turning points we set f prime(x) = 0
x2(8x + 15) = 0
bull This occurs when eitherbull x2 = 0 =rArr x = 0 (this is a point of inflection) More
bull 8x + 15 = 0 =rArr x = minus15
8= minus1875 (a minimum) More
bull To determine whether these are turning points are maximaminima or neither we find the second order derivative
f primeprime(x) =d
dxf prime(x) =
d
dx
(8x3 + 15x2
)= 24x2 + 30x
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Finding Turning Points
Example (f(x) = 2x4 + 5x3 continued)
bull The second order derivative is f primeprime(x) = 24x2 + 30x
bull We need to evaluate f primeprime(x) at the values of x we identified asturning points
bull f primeprime(0) = 0 =rArr a point of inflection More
bull f primeprime(minus1875) = 24times (minus1875)2 minus 30times 1875 = 28125 gt 0=rArr a minimum More
x
f(x)
-1875
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull
3x = 0 =rArr x = 0
orbull
6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) =
18 + 6x
Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull
f primeprime(0) = 18 gt 0 =rArr a minimum
bull
f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr
a minimum
bull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr
a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Your turn to find Maxima and Minima
Given f(x) = 9x2 + x3 previously you found f prime(x) = 18x + 3x2
bull To find the turning points set f prime(x) = 0
3x(6 + x) = 0
bull This occurs when eitherbull 3x = 0 =rArr x = 0 orbull 6 + x = 0 =rArr x = minus6
bull Second order derivative f primeprime(x) = 18 + 6x Evaluate this atthe possible turning points
bull f primeprime(0) = 18 gt 0 =rArr a minimumbull f primeprime(minus6) = 18 + 6times (minus6) = minus18 lt 0 =rArr a maximum
x
f(x)
-6
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w
U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points
2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w =
20
gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Maximising Utility
bull An investor gains what is known as utility from increasinghisher wealth (think of utility as simply enjoyment)
bull You can define someones utility as a function of wealth
Example (Your turn U(w) = 4w minus 110w2)
bull Differentiate U with respect to w U prime(w) = 4minus 2
10w
bull Set U prime(w) = 0 to find the critical points2
10w = 4 =rArr w = 20
bull w = 20 gives the theoretical level of wealth for this investorthat will maximise their utility
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Applications in Business
bull The ubiquitous Cobb-Douglas production function usesexponentials and logs More
bull The formal interpretation of regression coefficients ineconometrics requires differentiation More
bull Differentiation to finding maxima is used for constrainedoptimisation in operations management More
bull Marginal benefits and marginal costs can be derived usingdifferentiation More
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Summary
bull Functions log and exponential functions
bull Differentiation tells us about the behaviour of the function
bull The derivative of a single variable function is the tangent
bull The derivative can be interpreted as the ldquorate of changerdquo ofthe function
bull Chain rule product rule quotient rule
bull Finding maxima and minima
bull Applications in Business
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Summary of Differentiation Identities
Function Derivative
f(x) = axn f prime(x) = anxnminus1
f(x) = a (some constant) f prime(x) = 0
f(x) = expg(x) f prime(x) = gprime(x) middot expg(x)
y = logf(x) dy
dx=
f prime(x)
f(x)
y = f(u) u = g(x)dy
dx=
dy
dumiddot dudx
y = uv u = g(x) v = h(x)dy
dx= u middot dv
dx+ v middot du
dx
y =u
v u = g(x) v = h(x) dy
dx=
v middot dudxminus u middot dv
dxv2
More
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Additional Resources
bull Test your knowledge at the University of Sydney BusinessSchool MathQuizhttpquizeconusydeduaumathquiz
bull Additional resources on the Maths in Business websitesydneyeduaubusinesslearningstudentsmaths
bull The University of Sydney Mathematics Learning Centre has anumber of additional resources
bull Maths Learning Centre algebra workshop notes More
bull Other Maths Learning Centre Resources More
bull The Department of Mathematical Sciences and theMathematics Learning Support Centre at LoughboroughUniversity have a fantastic website full of resources More
bull Therersquos also tonnes of theory worked questions and additionalpractice questions online All you need to do is Google thetopic you need more practice with More
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
Differentiation Rules Application Conclusion
Acknowledgements
bull Presenters and content contributors Garth Tarr EdwardDeng Donna Zhou Justin Wang Fayzan Bahktiar PriyankaGoonetilleke
bull Mathematics Workshops Project Manager Jessica Morr fromthe Learning and Teaching in Business
bull Valuable comments feedback and support from Erick Li andMichele Scoufis
bull Questions comments feedback Let us know atbusinessmathssydneyeduau
